Enough Information?

$\begin{aligned}b&=(6-a) \quad \text{(from 1st equation)} \\ a^2+(6-a)^2+c^2&=18 \quad \text{(subst. into 2nd eqn.)} \\ 2a^2-12a+18+c^2&=0 \quad \text{(simplifying)} \\ 2(a-3)^2+c^2&=0 \quad \text{(factoring 1st 3 terms)}\end{aligned} \\$ The only way the sum of two squares can be zero is if both squares are zero. $\therefore a=3, c=0$ and $a-c=\boxed{3}$

We find that by arithmetic mean - geometric mean , the minimum value of $a^2+b^2$ when $a+b=6$ is 18, for which $a = b = 3$ . Plugging this into the other equation we get $18+c=18$ , for which $c = 0$ . Thus $a - c = 3 - 0 = 3$ .

By Cauchy-Schwarz Inequality , $\begin{aligned} (a^2+b^2)(1^2+1^2)&\geq (a+b)^2\\ \implies a^2+b^2&\geq 18\\ \implies a^2+b^2+c^2&\geq 18+c^2\\ \implies 18&\geq 18+c^2 \end{aligned}$

Hence $c^2\leq 0\implies c=0$ .Hence we're left with the system: $\begin{cases} a+b =6\\ a^2+b^2=18 \end{cases}$

Solving this system gives $a=b=3$ ,hence $a-c=3-0=\boxed{3}$

$b = 6 - a$

so

$a^2 + (6 - a)^2 + c^2 = 18$

simplify

$2a^2 - 12a + 36 + c^2 = 18$

solve for c

$c = \sqrt{-2a^2 + 12a - 18}$

simplify

$c = \sqrt{-2}(a - 3)$

$\sqrt{-2}$ will always be an imaginary number so the only real number solution for c is 0 where a = 3.

so

$a + c = 3 + 0 = 3$

Eq #1: Possible solutions (a,b) = (1,5), (5,1), (2,4), (4,2), (3,3)

Eq #2: (a,b,c) = (3,3,0),

Eq #3: a - c = 3 - 0 = 3

The question is symmetric in a and b. Therefore, as we know there is a single answer, a-c must equal b-c, and therefore a=b. Therefore a=b=3 and c=0.

5 squared = 25, which is more than 18, so, assuming whole numbers, the highest possible unknown value is 4

4 squared = 16, but no pair of squared numbers adds up to 2, which is needed for a sum of 18

2 squared = 4. Even if all three values are 2, the squares won't add up to 18

Therefore, at least one of the values must be 3

3 squared = 9. If two of the values are 3, we have our total of 18

But only if the third value is 0

The two values that are 3 must be a and b to have a sum of 6

Therefore, a = 3, b = 3, and c = 0

And a - c = 3 - 0 = 3

If we allow c=0 then by inspection a = 3 and b = 3 is a solution.

Nothing could be negative because squaring it would make it positive. So a and b needs to be 3. 3 squared + 3 squared is 18 so c had to be 0.
P.S. as a person who hated word problems growing up I think it is cheating to make two variables the same number. a and b are different letters and should be different numbers. This is a trick question for many students.

$a$ + $b$ = $6$ $[1]$

$a^2$ + $b^2$ + $c^2$ = $18$ $[2]$

Squaring $[1]:$

$a^2$ + $b^2$ = $36$ - $2ab$

Substituting into $[2]:$

$2a^2$ - $12a$ + [ $c^2$ + $18$ ]= 0

Solving for $a:$

$a$ = $3$ $\pm$ $i$ $\frac{\sqrt{2}}{2}$ $c$

Substituting $a$ into $b:$

$b$ = 3 $\mp$ $i$ $\frac{\sqrt{2}}{2}$ $c$

Since $a$ , $b$ and $c$ are real

$a = b = 3$ and $c = 0$

Thus $a - c = 3$

As the value of a-c is same for all value you can choose any two values of a and b which satisfy the first equation and find c. However it would be good if anyone posts any general solution.

I would do it slightly different.

Taking the constraint a,b,c>=0 . I rewrite eqn to get c^2=18-(a^2+b^2) From eqn 1 u know a^2+b^2= 36 -2ab So c^2=2ab-18

Now since c>=0 so c^2>=0

So the turning point is c^2=0. That would mean 18 -2ab =0 ab =9 . We know from eqn 1 a+b = 6

The only real solution to this is a=b=3 . This would give c= 0

Hence at our required turning point c=0 giving a-c = 3

Substitute b=6-a. The function y=a^2+(6-a)^2 has a parabolic graph, and reaches minimum of 18 when a=3. Since the right side of the second equation is precisely 18, there is a unique solution a-c=3. If it were <18 there could be no solution; if >18, infinitely many.

• If $a+b=6,$
• Then the minimum value of $a^2+b^2$ is:
• $a=b=\frac{a+b}{2}$ which gives $a=b=3$ .
• Therefore, this minimum value is calculated as $a^2+b^2=3^2+3^2=18$
• Then, we can easily find that $c^2=0$ , which gives us $c=0$ .
• The result of $a-c$ is:
• $a-c=3-0=0$ .

the only way to make 18 out of three square numbers is using zero squared plus a combination of three and minus three - the only way to make six from two of these is to use two threes - hence a and b both equal three and c equals zero

Draw the line a+b=6 and the sphere $a^2+b^2+c^2=18$ on coordinate axes a, b and c. They touch in only one place, at the point a=b=3, c=0 (radius is $\sqrt{18}$ and the length of the perpendicular from the line to the origin is also $\sqrt{18}$ by pythogoras). This point is the only one for which the two statements can both be true.

I just did it in my head: the only way to arrive at 18 is for a to be 4 and b and c to both be 1, 4-1 = 3. I now see that this is wrong because although this solution satisfies the second equation it does not work with the first. back to the drawing board. If a and b are both 3 this satisfies equation 1 and for equation 2 to work c must = 0, 3 - 0 =3. finito

a=3; b=3; a2+b2 is 9+9=18; c=0; 3-0 =3

The minimum value of a^2+b^2 subject to a+b=const is given by a=b=const/2. Hence, in this case a=b=3 returns the minimum value of a^2+b^2 which is equal 3^2+3^2=18! Then C^2 must be zero and consequently a-c=3-0=3.

Assuming a,b and c are WHOLE numbers (I speak English), not "real" numbers, and assuming they do not have negative values, then a+b=6 means both a and b could be 0,1,2,3,4, 5 or 6

However 5 or 6 squared exceed 18

4 squared is 16, but 1 squared is 1, 2 squared is 4 so cant add to 16 to make 18

Therefore only a and b are 3, 3 squared is 9

thus 9 +9 +0 = 18

So c = 0, simples ....

only 4 options for a+b =6 (0,6; 1,5; 2,4; 3,3). Squaring the two values leaves only 3,3 as an options. All others add up to value > 18. a and b being 3 gives a a + b b = 18, so c must be 0. 3-0 = 3.

(a+b)^2 = 6^2 a^2+2ab+b^2=36 -a^2-b^2-c^2=-18

2ab-c^2=18 a=6-b , b=6-a 2(6-b)b-c^2 = -18 = 2(6-a)a-c^2 12b-2b^2-c^2=12a-2a^2-c^2 12b-2b^2=12a-2a^2 --> a=b --> a=3 , b=3 9+9+c^2=18 --> c=0

a-c = 3

gut feeling

I'm no mathematician. The second equation only works if 2 of the numbers are 3 or -3, and the other 0. No other real numbers fit. Equation 1 only works if they are both 3. So a=b=3. So c has to be 0

I looked at all possible pairs of integers as values for a and b: (1,5), (2,4), (3,3), (4,2), (5,1).

Evaluating them, I realized that in 4 out of the 5 pairs, a^2 + b^2 was always greater than 18, and that would make c^2 a negative number, which is impossible; see below.

For example:

(When a=1, b=5): 1 + 25 + c^2 = 18? No. 26 plus anything would be > 18, making c^2 negative. Impossible.

(When a=2, b=4): 4 + 16 + c^2 = 18? No, 20 plus anything would be > 18, making c^2 negative. Impossible.

(When a=3, b=3): 9 + 9 + c^2 = 18? This could work...

(When a=4, b=2): 16 + 4 + c^2 = 18? No, 20 plus anything would be > 18, making c^2 negative. Impossible.

(When a=5, b=1): 25 + 1 + c^2= 18? No, 26 plus anything would be > 18, making c^2 negative. Impossible.

The one pair not greater than 18 was c^2 + c^2 = 18, making the correct pair (3,3). That made c = 0.

With the values a=3, b=3, c=0 in place, the equation became 3^2 + 3^2 + 0^2 = 9+9+0 = 18.

That meant a-c = 3-0 = 3.

a,b = 3; c=0 works (by inspection) So if the statement is generally true (which is implied by the question) then the answer is 3.